A353264 - OEIS

%I #5 Apr 09 2022 06:35:13

%S 1,4,15,64,48,60,144,16384,240,1300,1296,960,1008,3564,3840,

%T 1073741824,6000,14580,7056,20800,11520,25500,944784,245760,13104,

%U 24948,34560,57024,750000,16380,156816,4611686018427387904,102000,364500,46800,233280,134064,174636

%N a(n) is the least number k such that A018804(k)/k = n.

%C a(n) exist for all n>=1.

%C The solution k to A018804(k)/k = n is unique if and only if n is a power of 2: a(2^m) = 2^(2^(m+1)-2) = A051191(m+1).

%C The questions of existence and uniqueness are a part of a problem that was proposed during the Forty-Fifth International Mathematical Olympiad in Athens, Greece, July 7-19, 2004.

%D Dušan Djukić, Vladimir Janković, Ivan Matić, and Nikola Petrović, The IMO Compendium, A Collection of Problems Suggested for the International Mathematical Olympiads: 1959-2004, Springer, New York, 2006. See Problem 25, pp. 331, 726-727.

%H Dušan Djukić, Vladimir Janković, Ivan Matić, and Nikola Petrović, <a href="https://www.imomath.com/imocomp/sl04.pdf">IMO Shortlist 2004</a>, 2005, Problem 25, pp. 5-6, 18-19.

%H R. E. Woodrow, <a href="https://cms.math.ca/publications/crux/issue?volume=34&amp;issue=7">Problem N2</a>, The Olympiad Corner, Crux Mathematicorum, Vol. 34, No. 7 (2008), pp. 419-421.

%H <a href="/index/O#Olympiads">Index to sequences related to Olympiads</a>.

%e a(2) = 4 since A018804(4)/4 = 8/4 = 2 and 4 is the least number with this property.

%t f[p_, e_] := (e*(p - 1)/p + 1); r[n_] := Times @@ (f @@@ FactorInteger[n]); a[n_] := Module[{k = 1}, While[r[k] != n, k++]; k]; Array[a, 15]

%Y Cf. A018804, A051191.

%Y Subsequence of A066862.

%K nonn

%O 1,2

%A _Amiram Eldar_, Apr 09 2022